Problem: Let $f(x, y, z) = xy - x\sqrt{z}$ and $g(t) = (2, -t, 4t^2)$. $h(t) = f(g(t))$ $h'(1) = $
Explanation: Formula The multivariable chain rule says that $\dfrac{dh}{dt} = \nabla f(g(t)) \cdot g'(t)$. The $g'(t)$ part is how much a change in $t$ will cause the input to $f$ to move, and the $\nabla f(g(t))$ part is how much $f$ will change in response to this update to its input. [What's the intuition behind the formula?] Applying the formula We want to find $h'(1) = \nabla f(g(1)) \cdot g'(1)$. $\begin{aligned} &g(1) = (2, -1, 4) \\ \\ &g'(1) = (0, -1, 8t) = (0, -1, 8) \\ \\ &\nabla f = \left( y - \sqrt{z}, x, \dfrac{-x}{2\sqrt{z}} \right) \\ \\ &\nabla f(g(1)) = \nabla f(2, -1, 4) = \left( -3, 2, \dfrac{-1}{2} \right) \end{aligned}$ Substituting: $\begin{aligned} h'(1) &= \left( -3, 2, \dfrac{-1}{2} \right) \cdot (0, -1, 8) \\ \\ &= -2 - 4 \\ \\ &= -6 \end{aligned}$ Answer $h'(1) = -6$